Prime ideals equivalence

Question

I'm reading Herstein's Non-commutative Rings book. His definition of prime ring is

A ring $R$ is said to be a prime ring if $aAb=(0)$ then either $a=0$ or $b=0$.

Then he presents the following Lemma:

A ring $R$ is prime if and only if:

1. the right annihilator of a nonzero right ideal of $R$ must be $(0)$.
2. the left anninhilator of a nonzero left ideal of $R$ must be $(0)$.
3. if $A$, $B$ are ideals of $R$ and $AB=(0)$ then either $A=(0)$ or $B=(0)$.

I can prove that if $R$ is prime then (3) is true. Moreover, it's easy to see that (3) implies (1) and (2). I would like to prove that (1) or (2) implies that $R$ is prime. That's my attempt:

Suppose that (1) is true. Let $a,b \in R$ such that $aAb=(0)$. Assume that $a\not=0$. We know that $aA$ is a right ideal and $b$ belongs to the right annihilator of $aA$. If $aA\not=(0)$ then $b$ must be equal to $0$. But how can I show that $aA\not=0$? Is it even true for the case without unit in $R$?

Answer 1

It's also easy to see that $1$ or $2$ implies $3$, so we can consider all three equivalent and compare them to the initial definition of prime.

Suppose $A$ is a rng satisfying (any, hence all of) $1,2,3$, and that $aAb=\{0\}$. We aim to show either $a=0$ or $b=0$.

If $a=0$ we are done, so we consider when $a\neq 0$. By virtue of $1$, it must be that the right ideal $aA\neq \{0\}$. By virtue of $2$, $(aA)b=\{0\}$ implies $b=0$.